1037 Magic Coupon (25)

The magic shop in Mars is offering some magic coupons. Each coupon has an integer N printed on it, meaning that when you use this coupon with a product, you may get N times the value of that product back! What is more, the shop also offers some bonus product for free. However, if you apply a coupon with a positive N to this bonus product, you will have to pay the shop N times the value of the bonus product... but hey, magically, they have some coupons with negative N's!

For example, given a set of coupons {1 2 4 -1}, and a set of product values {7 6 -2 -3} (in Mars dollars M\$) where a negative value corresponds to a bonus product. You can apply coupon 3 (with N being 4) to product 1 (with value M\$7) to get M\$28 back; coupon 2 to product 2 to get M\$12 back; and coupon 4 to product 4 to get M\$3 back. On the other hand, if you apply coupon 3 to product 4, you will have to pay M\$12 to the shop.

Each coupon and each product may be selected at most once. Your task is to get as much money back as possible.

Input Specification:

Each input file contains one test case. For each case, the first line contains the number of coupons NC, followed by a line with NC coupon integers. Then the next line contains the number of products NP, followed by a line with NP product values. Here 1<= NC, NP <= 10^5^, and it is guaranteed that all the numbers will not exceed 2^30^.

Output Specification:

For each test case, simply print in a line the maximum amount of money you can get back.

Sample Input:

4
1 2 4 -1
4
7 6 -2 -3

Sample Output:

43

输入最多可以达到100000项，用cin输入会有一个测试点超时，用scanf()就行了；

 1 #include<iostream>
 2 #include<vector>
 3 #include<algorithm>
 4 using namespace std;
 5 int main(){
 6   int n, m, i, j;
 7   vector<int> a, b;
 8   cin>>n;
 9   a.resize(n);
10   for(i=0; i<n; i++) scanf("%d", &a[i]);
11   cin>>m;
12   b.resize(m);
13   for(i=0; i<m; i++) scanf("%d", &b[i]);
14   sort(a.begin(), a.end());
15   sort(b.begin(), b.end());
16   long sum=0;
17   for(i=0; a[i]<0 && b[i]<0; i++) sum += a[i]*b[i];
18   for(i=n-1, j=m-1; a[i]>0 && b[j]>0; i--, j--) sum += a[i]*b[j];
19   cout<<sum;
20   return 0;
21 }